Part CCSIR NET December 2024no-mean-and-still-a-weak-law

No mean and still a weak law

Let $X_{1}, X_{2}, \dots, X_{n}$ be independent and identically distributed (i.i.d.) random variables such that for $x = 3, 4, \dots, P (X_{1} = \pm x) = 1/ (2$ cx $^{2} ln (x))$ , where $c = \sum (x = 3..\infty) 1/ (x^{2} ln (x))$ . Then, which of the following statements are true?

A.E(| $X_{1}$ | $) = \infty$ ✓
B. $(1/ n) \sum (i = 1.. n) X_{i} \to 0$ in probability as $n \to \infty$ ✓
C.E(1/| $X_{1}$ | $) < \infty$ ✓
D. $E (X_{1}^{2}) = \infty$ ✓

You have the answer. Trap Analysis is why the other three were written.

Not a worked solution repeated four times — the specific reasoning error each wrong option was built to reward.

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No mean and still a weak law

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